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  1. #1
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    help with normal distribution question

    hi, i'm having trouble understanding this normal distribution question. any help on how would you work this out.thanks

    The present air-traffic volume at the City airport (number of takeoffs and landings) during the peak period has a normal distribution with a mean, μ, of 100 planes and a standard deviation, σ, of 20 planes

    (a) If the present runway capacity (for takeoffs and landings) is 142 planes per hour, what is the current probability of air traffic congestion in the peak period? (that is, what is the probability that the volume exceeds capacity - assume just one peak period daily).

    (b) If the peak period volume increases by 70 % over the next 10 years (that is, the peak period volume still has a normal distribution but the mean and standard deviation are both 70 % higher), what is the probability of peak period congestion if the runway capacity is unchanged from the present?

    (c) If the projected growth is correct, what airport capacity is required 10 years from now if the current service level is to be maintained (that is the same probability of peak period congestion as at present.
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    Quote Originally Posted by Tony213 View Post
    hi, i'm having trouble understanding this normal distribution question. any help on how would you work this out.thanks

    The present air-traffic volume at the City airport (number of takeoffs and landings) during the peak period has a normal distribution with a mean, μ, of 100 planes and a standard deviation, σ, of 20 planes

    (a) If the present runway capacity (for takeoffs and landings) is 142 planes per hour, what is the current probability of air traffic congestion in the peak period? (that is, what is the probability that the volume exceeds capacity - assume just one peak period daily).

    (b) If the peak period volume increases by 70 % over the next 10 years (that is, the peak period volume still has a normal distribution but the mean and standard deviation are both 70 % higher), what is the probability of peak period congestion if the runway capacity is unchanged from the present?

    (c) If the projected growth is correct, what airport capacity is required 10 years from now if the current service level is to be maintained (that is the same probability of peak period congestion as at present.
    a) Let X be the random variable present air-traffic volume. I may be wrong but it looks to me like you should calculate Pr(X > 142).


    Let Y be the random variable air-traffic volume in 10 years time. I may be wrong but it looks to me like you should:

    b) calculate Pr(Y > 142).

    c) find the value of a such that Pr(Y > a) = p, where p is the answer to a).
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    Quote Originally Posted by mr fantastic View Post
    a) Let X be the random variable present air-traffic volume. I may be wrong but it looks to me like you should calculate Pr(X > 142).

    for a what i did was 142-100/ 20
    = 2.1
    =.5-.4821
    =0.0179

    Let Y be the random variable air-traffic volume in 10 years time. I may be wrong but it looks to me like you should:

    b) calculate Pr(Y > 142).

    for b
    170 -142 / 34
    =0.823
    =.5-.2939
    =0.2061

    c) find the value of a such that Pr(Y > a) = p, where p is the answer to a).
    do the first 2 parts look right.

    also how would i answer question c) im really stuck on that on

    thanks
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    Quote Originally Posted by Tony213 View Post
    do the first 2 parts look right.

    also how would i answer question c) im really stuck on that on

    thanks
    The answer to (a) is correct (within the accuracy of your tables) but (b) is not (you have made a very simple error which I leave for you to find). The setting out for both is diabolical.

    I've shown how to do c): Find the value of a such that Pr(Y > a) = 0.0179.

    To elaborate:

    1. Find the value z_a such that \Pr(Z > z_a) = 0.0179 where Z is the standard normal variable (you must have been taught how to do this - use your tables in reverse ....).

    2. Then z_a = \frac{a - 170}{34} \Rightarrow a = \, ....
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    Quote Originally Posted by mr fantastic View Post
    The answer to (a) is correct (within the accuracy of your tables) but (b) is not (you have made a very simple error which I leave for you to find). The setting out for both is diabolical.

    I've shown how to do c): Find the value of a such that Pr(Y > a) = 0.0179.

    To elaborate:

    1. Find the value z_a of z such that \Pr(Z > z_a) = 0.0179 (you must have been taught how to do this - use your tables in reverse ....).

    2. Then z_a = \frac{a - 170}{34} \Rightarrow a = \, ....
    Regarding (b), do you honestly expect Pr(Z > 0.823) to be less than 0.5? Bearing in mind that 0.823 > 0 .....
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